Chapter 10, Problem 6P

To determine

To Find: The specific gravity of the mixture.

Answer to Problem 6P

The specific gravity of the mixture is $0.888$

Explanation of Solution

Given info:

Volume of antifreeze, $V_{af}=5\text{ L}$

Specific gravity of antifreeze, $S{G}_{af}=0.80$

Volume of water, $V_w=4\text{ L}$

Formula used:

Write the expression for the specific gravity.

  $\text{Specific gravity }SG=\frac{{\rho }_{\text{other fluid}}}{{\rho }_{\text{water}}}$

Write the expression for the gravity of a liquid.

  $\rho \text{ = }\frac{\text{Mass}}{\text{Volume}}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{m}_{af}={\rho }_{af}\times V_{af}=\left(S{G}_{af}\times {\rho }_w\right)\times V_{af}\\ m_{af}=0.8\times 1000\times 5=4000\text{ kg}\\ S{G}_{\text{mixture}}=\frac{{\rho }_{\text{mixture}}}{{\rho }_{\text{water}}}\text{ = }\frac{{\left(\frac{m}{V}\right)}_{\text{mixture}}}{{\rho }_{\text{water}}}\end{array}$

Mass of mixture is summation of mass of antifreeze and water.

  $\begin{array}{l}S{G}_{\text{mixture}}=\frac{m_{af}+m_w}{{\rho }_{\text{water}}\times V_{\text{mixture}}}\text{= }\frac{\left(S{G}_{af}\times {\rho }_{\text{water}}\times V_{af}\right)+\left({\rho }_{\text{water}}\times V_w\right)}{{\rho }_{\text{water}}\times V_{\text{mixture}}}\\ S{G}_{\text{mixture}}=\frac{\left(S{G}_{af}\times V_{af}\right)+\left(V_w\right)}{V_{\text{mixture}}}=\frac{\left(0.8\times 5\right)+\left(4\right)}{9}=0.888\end{array}$

Conclusion:

Thus, the specific gravity of the mixture is $0.888$ .